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For a unital ring $R$, there is a homomorphism $f:\mathbb{Z}\rightarrow R$ and the kernel is an ideal of the form $n\mathbb{Z}$ for unique $n\in \mathbb{N}$, which we call the characteristic of the ring.

In a similar kind of way, for a group $G$, and $g\in G$ there is a homomorphism $f:\mathbb{Z}\rightarrow G$ sending $n$ to $g^n$. The kernel of this homomorphism is again of the form $n\mathbb{Z}$ for unique $n\in \mathbb{N}$ which we call the order of $g\in G$, except for when $n=0$ and then we say $g$ has infinite order.

Wouldn't it be better to say $g$ has order zero in this case, for consistency?

amWhy
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Joshua Tilley
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  • I think it would be better to use "characteristic $\infty$". After all, that gives the size of the additive cyclic subgroup $\langle 1 \rangle$. – Randall Mar 18 '20 at 13:59
  • The order of $g\in G$ is the infimum of ${n\in {\mathbb N}: g^n=1}$. Now, the infimum of an empty subset of ${\mathbb R}$ is usually declared to be infinity. One more thing: In English, $0$ is not an element of ${\mathbb N}$. – Moishe Kohan Mar 18 '20 at 14:01
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    For me $0\in \mathbb{N}$ – Joshua Tilley Mar 18 '20 at 14:03
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    @MoisheKohan Whether $0$ is a natural number is not a question of language, but of the whims of the particular mathematician writing, and often depends more on which choice is more convenient in the given context (unlike whether $0$ is a positive number, which does depend on language). – Tobias Kildetoft Mar 18 '20 at 14:06
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    I have voted to close this as "opinion-based". – Geoffrey Trang Mar 18 '20 at 14:10
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    Where is a good place to ask this kind of question if not here? – Joshua Tilley Mar 18 '20 at 14:18
  • @TobiasKildetoft: It s not matter of language but of mathematical tradition. In English mathematical tradition, natural numbers start with $1$, while in French, they start with $0$. Ditto for neighborhoods: In French they are not required to be open while in English they are. – Moishe Kohan Mar 18 '20 at 14:28
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    What you say makes sense, but it is only one view. The other view is that the order of $g\in G$ is the order of the cyclic subgroup $\langle g\rangle$. I see no reason why we should adopt either view over the other. – user1729 Mar 18 '20 at 14:28
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    @Moishe As someone who was UK-educated and has only ever worked in UK maths departments, I find that the reality of this tradition is closer to what Tobias describes (possibly because when you write maths in English you do not expect your readers to be native English speakers and so be aware of such traditions). – user1729 Mar 18 '20 at 14:34
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    "elements of finite order" would mean all elements, if this were adopted. – rschwieb Mar 18 '20 at 14:57
  • @rschwieb but we'd then use "torsion elements" instead, which is relatively established. – user1729 Mar 18 '20 at 14:58
  • @user1729 I'm not sure why or how one would eradicate a well-used term (in the context of $\mathbb Z$-modules) just to adopt this convention. – rschwieb Mar 18 '20 at 15:17
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    @rschwieb I agree :-) – user1729 Mar 18 '20 at 15:35
  • @rschwieb, concerning elements of finite order, this would become elements of non-zero order though – Joshua Tilley Mar 18 '20 at 17:53
  • @JoshuaTilley It’s an alternative... but not an improvement. – rschwieb Mar 18 '20 at 18:01

3 Answers3

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The nomenclature for an infinite order element of a group is not in terms of the kernel of the relevant homomorphism, but rather the order of the cyclic subgroup generated by the element. For rings, they went with the nonnegative generator of the kernel of the homomorphism $\mathbb Z\to R$, though this is not explicit in the usual definition (which is inconsistent, saying the characteristic is the minimal positive integer such that $n1=0$, then giving $0$ as an exception when no such integer exists).

One could argue that the subgroup concept makes more sense for general groups because it is given in terms of the order of a group, whereas the kernel definition is additive which is appropriate for the additive group of a ring.

Matt Samuel
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I think it's harmless to call such elements of order zero; probably it's been done occasionally. If for some reason it's practical (e.g., you talk of elements of order $n$ and $n$ can be zero), just say it at the beginning.

Actually it's "almost" done somewhere: the characteristic of a unital ring $R$ is by definition the order of $1$ in $(R,+)$... and one says "characteristic zero", and (almost?) never "infinite characteristic".

YCor
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Here's a situation where this convention is useful. Suppose we have a group homomorphism $f : G \to H$ and let $a$ be in $G$. What is the order of $f(a)$ in $H$? In general, $f$ need not preserve order (just consider the constant map $f(a) = 1$), but what's true is that $\operatorname{ord}_H(f(a))$ must be a divisor of $\operatorname{ord}_G(a)$. If we set $\operatorname{ord}(a) = 0$ for elements of "infinite" order, then, keeping in mind that everything divides 0 while 0 divides only 0, this divisibility relation shows that homomorphic images of elements of infinite order can have arbitrary order, while no element of finite order can map to an element of infinite order.

Unit
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